• LeetCode 443. String Compression


    原题链接在这里:https://leetcode.com/problems/string-compression/

    题目:

    Given an array of characters, compress it in-place.

    The length after compression must always be smaller than or equal to the original array.

    Every element of the array should be a character (not int) of length 1.

    After you are done modifying the input array in-place, return the new length of the array.

    Follow up:
    Could you solve it using only O(1) extra space?

    Example 1:

    Input:
    ["a","a","b","b","c","c","c"]
    
    Output:
    Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
    
    Explanation:
    "aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

    Example 2:

    Input:
    ["a"]
    
    Output:
    Return 1, and the first 1 characters of the input array should be: ["a"]
    
    Explanation:
    Nothing is replaced.

    Example 3:

    Input:
    ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
    
    Output:
    Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
    
    Explanation:
    Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
    Notice each digit has it's own entry in the array.

    题解:

    Accumlate the count of repeating chars, if count > 1, append to the char.

    Use (""+count).toCharArray() to easily append int as char array. 

    Time Complexity: O(chars.length). Space: O(1).

    AC Java:

     1 class Solution {
     2     public int compress(char[] chars) {
     3         if(chars == null || chars.length == 0){
     4             return 0;
     5         }
     6         
     7         int pos = 0;
     8         int count = 0;
     9         int i = 0;
    10         while(i<chars.length){
    11             char cur = chars[i];
    12             while(i<chars.length && chars[i] == cur){
    13                 count++;
    14                 i++;
    15             }
    16             
    17             chars[pos++] = chars[i-1];
    18             if(count > 1){
    19                 for(char c : (""+count).toCharArray()){
    20                     chars[pos++] = c;
    21                 }
    22             }
    23 
    24             count = 0;
    25         }
    26         
    27         return pos;
    28     }
    29 }

    类似Encode and Decode StringsCount and SayDesign Compressed String Iterator.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/8202915.html
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